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# Every one who passes the test will be awarded a degree. The probabilit

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Joined: 31 Oct 2013
Posts: 1473
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Every one who passes the test will be awarded a degree. The probabilit  [#permalink]

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11 Sep 2018, 23:06
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15% (low)

Question Stats:

80% (01:17) correct 20% (01:18) wrong based on 70 sessions

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Every one who passes the test will be awarded a degree. The probability that Tom passes the test is 0.3, and the the probability that Jhon passes the test is 0.4. The 2 events are independent of each other. What is the probability that at least one of them gets the degree?

A) 0.28
B) 0.32
C) 0.5
D) 0.58
E) 0.82
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Joined: 02 Sep 2009
Posts: 58320
Re: Every one who passes the test will be awarded a degree. The probabilit  [#permalink]

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11 Sep 2018, 23:15
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selim wrote:
Every one who passes the test will be awarded a degree. The probability that Tom passes the test is 0.3, and the the probability that Jhon passes the test is 0.4. The 2 events are independent of each other. What is the probability that at least one of them gets the degree?

A) 0.28
B) 0.32
C) 0.5
D) 0.58
E) 0.82

P(at least 1) = 1 - P(none) = 1 - 0.7*0.6 = 0.58.

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Joined: 15 Aug 2018
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GMAT 1: 740 Q47 V45
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Re: Every one who passes the test will be awarded a degree. The probabilit  [#permalink]

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23 Dec 2018, 06:16
Bunuel Why isn't it 0.82?

P(T) = 0.3
P(J) = 0.4

P(T or J) = 0.7 - 0.12 = 0.58

I completely understand your approach. But I am confused by the word at least.

Since at least, includes the cases that 1) P (T) happens with a probability of 0.3 2) happens with a probability of 0.4 and both happen with a probability of 0.12. I am not really getting the gist of why we are subtracting the 0.12 case rather than adding it... I assume that the overlap of 0.12 is included in 0.3 + 0.4, but I don't really understand why...

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Re: Every one who passes the test will be awarded a degree. The probabilit  [#permalink]

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23 Dec 2018, 06:28
1
OA:D

Probability of Tom passing P(TP): 0.3
Probability of Tom failing P(TF): 1-0.3=0.7

Probability of Jhon passing P(JP): 0.4
Probability of Jhon failing P(JF): 1-0.4=0.6

Probability that at least one pass : P(TP)*P(JF) + P(TF)*P(JP)+ P(TP)*P(JP) = 0.3*0.6+0.4*0.7+0.3*0.4=0.18+0.28+0.12=0.58
Re: Every one who passes the test will be awarded a degree. The probabilit   [#permalink] 23 Dec 2018, 06:28
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